Lecture 17 - Global Gauss Bonnet Theorem

#math

Theorem

Let $R \subset S$ be a regular region of an oriented surface and let $δ R$ be made up of $n$ closed piecewise simple regular curves, $c_{1}, . . ., c_{n}$ . Then:

\sum_{i}^{n} φ_{i} + \int_{R} K d A + \int_{δ R} K_{g} d s = 2 π χ

Where $χ$ is the Euler characteristic of the region and $φ_{i}$ are the external angles formed by the intersection of the curves.

Sketch of the proof

Let $R \subset S$ be a regular region of an oriented surface $S$ . Let $\partial R$ be the boundary of $R$ be made up by closed piecewise simple regular curves $c_{1}, . . ., c_{n}$ , and let $φ_{i}$ be the exterior angle formed by the intersection of these curves.
Let $T = {R_{i} : i = 1, . . ., F}$ , where $F$ is the number of faces, be a triangulation of $R$ for each $i$ . Let $e_{i j}, j = 1, 2, 3$ be the edges and let $ϑ_{i j}$ denote the exterior angle formed by the triangulation and $θ_{i j}$ denote the respective interior angle with the relation $θ_{i j} = π - ϑ_{i j}$ .
Using the local Gauss Bonett Theorem on each face we obtain:

\sum_{i = 1}^{F} (\int K d A + \sum_{j = 1}^{3} \int_{γ_{i j}} K_{g} d s + \sum_{j = 1}^{3} (π - θ_{i j})) = \sum_{i = 1}^{F} 2 π

The edges shared by faces are cancelled out during the integration:

\int_{R} K d A + \int_{δ R} K_{g} d s + \sum_{i, j} (π - θ_{i j}) = \sum_{i = 1}^{F} 2 π

I will now introduce the following notation:

$E_{e} =$ number of external edges of $T$
$E_{i}$ = the number of internal edges of $T$
$V_{e}$ = the number of external vertices of $T$
$V_{i}$ = the number of internal vertices of $T$

As an example let us look at the following triangulation of a region:
$Triangulation.png|500$
We have $E_{e} = 3, E_{i} = 3, V_{e} = 3, V_{i} = 1, F = 3$ . We convince ourselves of the following relations:

\begin{aligned} 3 F & = 2 E_{i} + E_{e} \\ E_{e} & = V_{e} \end{aligned}

Thus $\sum_{i j} (π - θ_{i j}) = 3 π F - \sum_{i j} θ_{i j} = 2 π E_{i} + π E_{e} - \sum_{i j} θ_{i j}$ .
The sum of the all the internal angles around an internal vertex is $2 π$ and if a vertex was introduced such that it lies on a curve (thus it is an external vertex) then the internal angles add up to $π$ . For clarity we will split $V_{e} = V_{e t} + V_{e c}$ where $V_{e t}$ denotes the vertices added during triangulation while $V_{e c}$ are the 'natural' vertices formed by the curves $c_{i}$ making up the boundary of the region. Naturally we have that the sum of the interior angles corresponding to $V_{e c}$ is equal to $\sum_{i = 1}^{n} (π - φ_{i})$ Thus we have that:

\begin{aligned} \sum_{i, j} θ_{i j} & = 2 π V_{i} + π V_{e t} + \sum_{i = 1}^{n} (π - φ_{i}) \\ = 2 π V_{i} + π V_{e t} + π V_{e c} - \sum_{k = 1}^{n} (φ_{k}) \\ = 2 π V_{i} + π V_{e} - \sum_{k = 1}^{n} (φ_{k}) \end{aligned}

Now using the fact $E_{e} = V_{e}$ and adding and subtracting $π E_{e}$ we obtain that:

\sum_{i j} (π - θ_{i j}) = 2 π E - 2 π V + \sum_{i = 1}^{n} φ_{i}

Thus finally:

\int_{R} K d A + \int_{δ R} K_{g} d s + \sum_{i = 1}^{n} φ_{i} = 2 π (F - E + V) = 2 π χ

Corollary

For a sphere $S$ if we take the boundary of our region $R$ to be a great circle then:

\int_{R} K = 2 π χ

Consider a smooth vector field $F (x, y) = (P (x, y), Q (x, y))$ on a subset $U$ of $R^{2}$ . Let $C (t) = (x (t), y (t))$ be a parameterized curve in $U$ from a point $p$ to $q$ with $a \leq t \leq b$ .
A necessary condition for the vector field to be the gradient to some scalar function $f$ is $P_{y} = Q_{x}$ .

We are interested in asking if $P_{y} - Q_{x} = 0$ holds for a vector field $F = (P, Q)$ on $U$ then is it the gradient of some scalar function $f (x, y)$ on $U$ ?

Vector Field	Differential Form
$F = (P, Q)$	$ω = P d x + Q d y$
$G r a d f = (f_{x}, f_{y})$	$d f = f_{x} d x + f_{y} d y$
$Q_{x} - P_{y} = 0$	$d ω = (Q_{x} - P_{y}) d x \land d y = 0$
If $ω = d f$ this is called the exact form.

Let $Z^{k} (M)$ be the vector space of all closed $k$ -forms and $B^{k} (M)$ be the vector space of all exact $k -$ forms. Then:

H^{k} (M) = Z^{k} (M) / B^{k} (M)

measures the extent to which closed $k -$ forms fail to be exact.